Past exam of the electromagnetism and optics course of the University of Oxford / 2016 / 7 / Solution

Write $\mu ={\mu }_0{\mu }_r$ and $ϵ={ϵ}_0{ϵ}_r$ for homogeneous scalar material constants, and use $\mathbf{J}=\sigma \mathbf{E}$. Taking the curl of Faraday's law of induction and substituting the Ampere-Maxwell law givesThe vector calculus identity for a double curl therefore yieldsThe second term represents electric current response, rather than displacement current. A dispersive $\sigma (\omega )$ is used in the frequency domain; its general time-domain counterpart is a temporal response convolution.

For the convention $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, the plane wave equation isDecompose the electric field into components parallel and perpendicular to $\mathbf{k}$. With $S=\mu ϵ{\omega }^2+i\mu \sigma \omega$, the longitudinal eigenvalue is $S$ and each transverse eigenvalue is $S-k^2$; thus the determinant is $S(S-k^2{)}^2$. For a transverse electromagnetic wave, $\mathbf{k}\cdot {\mathbf{E}}_0=0$ andwhen ${\mu }_r=1$ and the background ${ϵ}_r\simeq 1$. A real dissipative electrical conductivity generally makes $k$ complex, with the decaying branch chosen for a passive medium.

For a cold, collisionless, unmagnetized plasma, treat ions as stationary and write $e>0$ for the magnitude of the elementary charge. The electron equation of motion is $-i\omega m_e\mathbf{v}=-e\mathbf{E}$, soThis reactive response includes no collision-induced Joule heating. Substitution into the transverse dispersion relation givesThe quantity ${\omega }_p$ is the electron plasma frequency. The refractive index is real above this cutoff; below it, the wave is evanescent. The electron response is represented here by $\sigma (\omega )$, so it must not also be counted a second time in the background permittivity.

For a nonzero longitudinal component, taking the dot product of the matrix equation with $\mathbf{k}$ instead gives $S(\mathbf{k}\cdot {\mathbf{E}}_0)=0$. The same cold plasma response makes $S=({\omega }^2-{\omega }_p^2)/c^2$, henceAt nonzero $k$, the remaining equation forces ${\mathbf{E}}_0$ to be parallel to $\mathbf{k}$, and Faraday's law of induction gives ${\mathbf{B}}_0=0$. These are electrostatic Langmuir waves, with no $k$ dependence of angular frequency in the zero-temperature model. Equilibrium neutrality does not forbid their perturbed charge density; Gauss's law requires such a density when $\mathbf{k}\cdot {\mathbf{E}}_0\ne 0$.